UVa1605 Building for UN 构造

The United Nations has decided to build a new headquarters in Saint Petersburg, Russia. It will have a 
form of a rectangular parallelepiped and will consist of several rectangular floors, one on top of another. 
Each floor is a rectangular grid of the same dimensions, each cell of this grid is an office. 
Two offices are considered adjacent if they are located on the same floor and share a common wall, 
or if one’s floor is the other’s ceiling. 
The St. Petersburg building will host n national missions. Each country gets several offices that 
form a connected set. 
Moreover, modern political situation shows that countries might want to form secret coalitions. For 
that to be possible, each pair of countries must have at least one pair of adjacent offices, so that they 
can raise the wall or the ceiling they share to perform secret pair-wise negotiations just in case they 
need to. 
You are hired to design an appropriate building for the UN.

Input

Input consists of several datasets. Each of them has a single integer number n (1 ≤ n ≤ 50) — the 
number of countries that are hosted in the building.

Output

On the first line of the output for each dataset write three integer numbers h, w, and l — height, width 
and length of the building respectively. 
h descriptions of floors should follow. Each floor description consists of l lines with w characters on 
each line. Separate descriptions of adjacent floors with an empty line. 
Use capital and small Latin letters to denote offices of different countries. There should be at most 
1 000 000 offices in the building. Each office should be occupied by a country. There should be exactly 
n different countries in the building. In this problem the required building design always exists. 
Print a blank line between test cases.

Sample Input

4

Sample Output

2 2 2 
AB 
CC 
zz 
zz

有一种叫做中途相遇法的东西，也真是神奇了；只堆两层，如果看不懂，把代码运行一下，输入一个样例，就明白了！

代码如下：

//UVa1605 Building for UN
#include <stdio.h>
#include <string.h>
int up[52][52];
int down[52][52];
char c[55] ="aabcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
int main() {
	int n ;
	while(~scanf("%d",&n)) {
		printf("%d %d %d\n",2,n,n);
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				up[i][j] = i;
				down[j][i] = i;
			}
		}
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				printf("%c",c[up[i][j]]);
			}
			printf("\n");
		}
		printf("\n");
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				printf("%c",c[down[i][j]]);
			}
			printf("\n");
		}		
	}
	return 0;
}